Helen made the conjecture that "every multiple of six has more
factors than the two numbers either side of it". Is this conjecture
true?

Mini-max

Stage: 3 Challenge Level:

Thomas, James, Jonathan Liong, Scott Reynolds, Ren Papilion, and
Tom Stevens from St Peters College, Adelaide, Australia all
discovered that writing down the $90$ two digit numbers
($10,11,\ldots ,99$) uses nine $0$'s, and nineteen of each of the
digits $1,2,\ldots ,9$. To check this we see that $9+(19\times 9) =
180$, which gives the correct number of digits altogether.

Ling Xiang Ning, Tao Nan School, Singapore and Sheila Luk, Joyce Fu
and Sam Webster, Year 10, The Mount School, York all spotted the
rule for the number of occurrences of each digit in writing down
all $n$ digit numbers.

Elizabeth Hyde, age 13 from Haybridge High School, Hagley West
Midlands went on to derive this general formula using a spreadsheet
and M. Ali Khan, age 18, Reigate College also gave a clear
logically argued solution.

Solution for three digit
numbers: For three digits there are $900$ numbers
($1,2,\ldots ,999$ excluding $1,2,\ldots ,99$) and hence $2700$
digits.

Consider the occurrences of the digit $1$. There is only one number
in which this digit occurs three times, that is the number $111$.

Now consider numbers in which $1$ occurs exactly twice. There are
eight of these of the form $x11$ (where $x$ is some other digit and
$x \neq 0, 1$ and nine of the form $1x1$ (where $x\neq 1$), and
nine of the form $11x$ (where $x\neq 1$). This makes a total of
$52$ occurrences of the digit $1$ for these numbers.

Finally, consider the numbers in which $1$ occurs exactly once.
There are $81$ of these of the form $1xy$ (where $x$ and $y$ are
other digits and $x,y\neq 1$), $72$ of the form $x1y$ (where $x\neq
0,1$ and $y\neq 1$), and $72$ of the form $xy1$ (where $x\neq 0,1$
and $y\neq 1$). This makes a total of $180$ occurrences of the
digit $1$ for these numbers, and hence a grand total of $280$
occurrences of the digit $1$.

Now the same argument holds with $1$ replaced by any of the other
digits $2,3,\ldots ,9$ so the number of occurrences of $0$ is
$2700-(8\times 280)=180$. We can check this directly in two
different ways. First, we can count the number of occurrences of
$0$ in numbers of the forms $xy0$, $x0y$ and $x00$. Second, we
would expect there to be $100$ fewer $0$'s than any other digit
because none of the numbers can start with $0$.

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